Corollary 4.8.1.23. Let $n$ be an integer and let $F_0: \operatorname{\mathcal{C}}_0 \rightarrow \operatorname{\mathcal{C}}$ and $F_1: \operatorname{\mathcal{C}}_1 \rightarrow \operatorname{\mathcal{C}}$ be functors of $(n,1)$-categories. Then the oriented fiber product $\operatorname{\mathcal{C}}_0 \operatorname{\vec{\times }}_{\operatorname{\mathcal{C}}} \operatorname{\mathcal{C}}_1$ and the homotopy fiber product $\operatorname{\mathcal{C}}_0 \times ^{\mathrm{h}}_{\operatorname{\mathcal{C}}} \operatorname{\mathcal{C}}_1$ are also $(n,1)$-categories.
$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$
Proof. By definition, the oriented fiber product $\operatorname{\mathcal{C}}_0 \operatorname{\vec{\times }}_{\operatorname{\mathcal{C}}} \operatorname{\mathcal{C}}_1$ can be realized as an iterated fiber product
\[ \operatorname{\mathcal{C}}_0 \times _{ \operatorname{Fun}( \{ 0\} , \operatorname{\mathcal{C}}) } \operatorname{Fun}( \Delta ^1, \operatorname{\mathcal{C}}) \times _{ \operatorname{Fun}( \{ 1\} , \operatorname{\mathcal{C}}) } \operatorname{\mathcal{C}}_1, \]
which is an $(n,1)$-category by virtue of Propositions 4.8.1.22 and Remark 4.8.1.18. The homotopy fiber product $\operatorname{\mathcal{C}}_0 \times ^{\mathrm{h}}_{\operatorname{\mathcal{C}}} \operatorname{\mathcal{C}}_1$ is a full subcategory of $\operatorname{\mathcal{C}}_0 \operatorname{\vec{\times }}_{\operatorname{\mathcal{C}}} \operatorname{\mathcal{C}}_1$, which coincides with $\operatorname{\mathcal{C}}_0 \operatorname{\vec{\times }}_{\operatorname{\mathcal{C}}} \operatorname{\mathcal{C}}_1$ for $n \leq -1$. Applying Remark 4.8.1.16, we see that it is also an $(n,1)$-category. $\square$